Graphs of velocity versus time are shown below for a car and a motorcycle travelling along the same road. The car passes the stationary motorcycle at t = 0. http://i.imgur.com/rnXsd6M.jpg

What is the average acceleration (in km/h/s) of the motorcycle during the first 15 seconds? a = Δv/Δt = 100/15 = 6.67 m/s2 = 24km/h/s

At what time does the motorcycle overtake the car? I figured that the point at which their distances are equal is when the motorcycle overtakes the car, and as such did (using the average acceleration above): d(motorcycle) = 3.33t2 d(car) = vt = 22.22t 3.33t2 = 22.22t t = 6.7 seconds

January 8th 2014, 03:55 PM

romsek

Re: Physics - Linear Motion

Quote:

Originally Posted by Fratricide

Graphs of velocity versus time are shown below for a car and a motorcycle travelling along the same road. The car passes the stationary motorcycle at t = 0. http://i.imgur.com/rnXsd6M.jpg

What is the average acceleration (in km/h/s) of the motorcycle during the first 15 seconds? a = Δv/Δt = 100/15 = 6.67 m/s2 = 24km/h/s

At what time does the motorcycle overtake the car? I figured that the point at which their distances are equal is when the motorcycle overtakes the car, and as such did (using the average acceleration above): d(motorcycle) = 3.33t2 d(car) = vt = 22.22t 3.33t2 = 22.22t t = 6.7 seconds

You can't use the average acceleration. You have to use the actual velocity as shown on the graph.
The distance each vehicle travels is the integral of the velocities shown on the graph w/respect to time.
So compute those two integrals and determine where they are equal.

You're going to have to split the motorcycle integral into pieces, for t = [0,10), [10,15), [15,40) since the velocity changes form across these intervals.

January 9th 2014, 10:50 AM

ebaines

Re: Physics - Linear Motion

You must also be careful with units. I strongly suggest that you convert velocity to m/s and acceleration to m/s^2 - the acceleration of the motorcycle in the first phase is 80 KPH/10s x 1000 m/Km x 1/3600 Hr/s = 2.22 m/s^2, and in the second phase is 20 KPH/5s = 1.11 m/s^2. Finally, when you get an answer think about whether it makes sense - your answer of 6.7 seconds means the motorcycle catches up to the car when its only going about 60 KPH, so it's not even going as fast as the car yet. Clearly the answer must be greater than 10 seconds.

January 9th 2014, 02:59 PM

Fratricide

Re: Physics - Linear Motion

I am unsure as to how to use the different accelerations from the intervals of the motorbike's motion in order to find the integral I need to equate to the integral of the car.

January 9th 2014, 05:27 PM

romsek

Re: Physics - Linear Motion

Quote:

Originally Posted by Fratricide

I am unsure as to how to use the different accelerations from the intervals of the motorbike's motion in order to find the integral I need to equate to the integral of the car.

eh you know how to do it.

Take a look at the first interval.

It's linear from 0-10s with slope 8 km/hr s. This slope is the acceleration during this interval.

These units are going to cause trouble so we can convert this to a acceleration of 20/9 m/s^2

so you can say in this interval

m/s

this integrates easily enough, and noting that pbike(0)=pcar(0)=0 we get

m

Evaluating this at t=10 we find

m

The car has been moving at a constant speed of 80 km/hr during this time.

Converting this speed to m/s we get

m/s so

m

so at t=10 the position of the car is

m

so the bike hasn't caught up by 10s.

Repeat this process for the second interval but now the acceleration is different. If by 15s the bike still hasn't caught up repeat it for the third interval that now has constant velocity.

It's just a couple of simple integrals and some algebra.

position is the integral of velocity
velocity is the integral of acceleration

see if you can work it now.

January 10th 2014, 04:45 AM

ebaines

Re: Physics - Linear Motion

I'm guessing that the OP is taking an introductory physics class and so does not know what an integral is. Let's keep it simple - the distance covvered is the area under the velocity curve. Note that from t=0 to t-10 s the velocity curve forms a triangle. Since the area of a triangle is one half base times height, the distance travelled from t=0 to t= 10 seconds is (1/2)(10s)(80Km/Hr). Do the conversion from Km/Hr to m/s, multiply this out and you get 111.1 m. For the second phase from t=10 to t=15s the area under the velocity graph is a trapezoid, and the area of that is (1/2)(80 Km/Hr+100Km/Hr)(5 s). Again, convert from Km/Hr to m/s and you get the area is 125m. Finally for t>15s the area is simply a rectangle, whose area as a function of time is 100 Km/Hr x (t-15s). So, for t>15 the total distamnce travelled by the motorcycle is 111.1m + 125m + 27.8m/s(t-15s). Set that equal to the area under the car's velocity curve, which is a simple rectangle and is t times 22.2m/s. Now solve for t.